Q.What are the differences between coulomb force and gravitational force ?

Inverse Square Law Comparison — Why the Formula Holds

The Inverse Square Law appears in physics wherever a quantity spreads out uniformly from a point source in three-dimensional space. The core idea is that the intensity (or field strength) decreases as the square of the distance from the source.

1. The Intuition: Spreading Over a Sphere

Imagine a point source emitting energy, light, sound, or gravitational force equally in all directions.

• At a distance $r$, the energy is spread uniformly over the surface area of a sphere of radius $r$.
• The surface area of a sphere is:

If the total power (or flux) emitted by the source is $P$, then the intensity $I$ (power per unit area) at distance $r$ is:

Key insight: The same total power is spread over a larger and larger area as $r$ increases. Hence, intensity is inversely proportional to $r^2$.

2. Derivation for Gravitational Force (Newton's Law)

Newton’s law of gravitation states:

Why $1/r^2$?

• The gravitational field lines from a point mass $M$ radiate outward uniformly.
• The number of field lines crossing a sphere of radius $r$ is constant (conservation of flux).
• The density of field lines (force per unit mass) at distance $r$ is:

• This is because the total flux $\Phi = 4\pi G M$ is spread over $4\pi r^2$, giving:

Thus, the force on a test mass $m$ is $F = m g = \frac{G M m}{r^2}$.

3. Derivation for Coulomb's Law (Electrostatics)

Coulomb’s law for electric force between two point charges $q_1$ and $q_2$:

Why $1/r^2$?

• Electric field lines from a point charge $q$ radiate radially outward (or inward for negative charge).
• Gauss’s law states that the total electric flux through a closed surface is proportional to the enclosed charge:

• For a sphere of radius $r$ centered on the charge, the field is radial and constant in magnitude:

• Therefore:

• The force on a test charge $q_2$ is $F = q_2 E = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$.

4. Derivation for Light/Radiation Intensity

For a point source of light emitting power $P$:

• At distance $r$, the power is spread over a sphere of area $4\pi r^2$.
• Illuminance (intensity) is:

Why not $1/r$?

• In 2D (e.g., a line source), intensity falls as $1/r$ because the circumference of a circle is $2\pi r$.
• In 3D, the surface area grows as $r^2$, so intensity falls as $1/r^2$.

5. The Common Mathematical Reason

All inverse square laws arise from conservation of flux in three-dimensional space with isotropic emission. The geometry forces: …